Integration (1 Viewer)

[taggs-sasuke]

I would really appreciate help with these two problems. Thanks!

1)

The region bounded by the curve y = x^(1/2) - 1, the lines x=2, x=3, and the x-axis, is rotated about the x-axis. Find the volume of the solid of revolution so formed.

Answer: [ 7/2 - 4.3^(1/2) +(8/3).2^(1.2) ] units^3

2)

I can't start this particular question because I don't know how to sketch

y = 1 / [ (1+x)^(1/2) ].

 

[tommykins]

cbf doing 1 but to do 2 you simply draw sqrt(1+x) and do the recipricol

 

[gurmies]

 

[taggs-sasuke]

tommykins:

I can sketch y = (1+x)^(1/2) but I don't know how to sketch the reciprocal (in fact, I don't know how to sketch the reciprocal of any function TT).

gurmies:

Thanks! I also realised the answer I typed out is missing pie.

 

[tommykins]

assymtote at whenever the graph has a root.

you then use logic for the shape of the curve.

 

[taggs-sasuke]

assymtote at whenever the graph has a root.

you then use logic for the shape of the curve.

Oh, ok. Thanks. I'll give it a try.

 

[shaon0]

Someone's learnt integration.

 

[bored of sc]

tommykins:

I can sketch y = (1+x)^(1/2) but I don't know how to sketch the reciprocal (in fact, I don't know how to sketch the reciprocal of any function TT).

Also, values of x for when graph is positive/negative is the same i.e. since graph is postive for all values of x (I think) than the reciprocal graph will also be positive for all values of x.